Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 55, L = 20;
int n, m, q, nv[N], pai[N], logn, anc[N][L];
vector<int> minimum10[N][L], c[N];
std::vector<int> g[N];
void dfs(int x, int p) {
  for (int i = 0; i < (int)g[x].size(); i++) {
    int y = g[x][i];
    if (y != p) {
      nv[y] = nv[x] + 1;
      pai[y] = x;
      dfs(y, x);
    }
  }
}
std::vector<int> join(vector<int> a, vector<int> b) {
  std::vector<int> c;
  sort(a.begin(), a.end());
  sort(b.begin(), b.end());
  c.resize((int)a.size() + (int)b.size());
  merge(a.begin(), a.end(), b.begin(), b.end(), c.begin());
  sort(c.begin(), c.end());
  c.resize(distance(c.begin(), unique(c.begin(), c.end())));
  if ((int)c.size() > 10) c.resize(10);
  return c;
}
void build() {
  logn = 31 - __builtin_clz(n);
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= logn; j++) {
      anc[i][j] = -1;
    }
  }
  for (int i = 1; i <= n; i++) {
    anc[i][0] = pai[i];
    for (auto x : c[i]) minimum10[i][0].push_back(x);
  }
  for (int j = 1; j <= logn; j++) {
    for (int i = 1; i <= n; i++) {
      if (anc[i][j - 1] == -1) continue;
      anc[i][j] = anc[anc[i][j - 1]][j - 1];
      minimum10[i][j] =
          join(minimum10[i][j - 1], minimum10[anc[i][j - 1]][j - 1]);
    }
  }
}
std::vector<int> query(int a, int b) {
  if (nv[a] < nv[b]) swap(a, b);
  vector<int> ans;
  ans = join(ans, minimum10[a][0]);
  for (int j = logn; j >= 0; j--) {
    if (nv[a] - (1 << j) >= nv[b]) {
      ans = join(ans, minimum10[a][j]);
      a = anc[a][j];
    }
  }
  if (a == b) {
    ans = join(ans, minimum10[a][0]);
    return ans;
  }
  ans = join(ans, minimum10[b][0]);
  for (int j = logn; j >= 0; j--) {
    if (anc[a][j] != -1 && anc[a][j] != anc[b][j]) {
      ans = join(ans, minimum10[a][j]);
      a = anc[a][j];
      ans = join(ans, minimum10[b][j]);
      b = anc[b][j];
    }
  }
  ans = join(ans, minimum10[a][0]);
  ans = join(ans, minimum10[b][0]);
  ans = join(ans, minimum10[anc[a][0]][0]);
  ans = join(ans, minimum10[anc[b][0]][0]);
  return ans;
}
int main() {
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (int i = 1; i <= m; i++) {
    int x;
    scanf("%d", &x);
    c[x].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    sort(c[i].begin(), c[i].end());
    if (c[i].size() > 10) c[i].resize(10);
  }
  dfs(1, 0);
  build();
  while (q--) {
    int a, b, k;
    scanf("%d %d %d", &a, &b, &k);
    std::vector<int> ans = query(a, b);
    printf("%d ", min(k, (int)ans.size()));
    for (int i = 0; i < min(k, (int)ans.size()); i++) {
      printf("%d ", ans[i]);
    }
    printf("\n");
  }
  return 0;
}